Question

In the process shown in the figure for an ideal diatomic gas, the value of q and $\Delta H$ respectively are:

• A. $79.5P_1{V}_1\text{and}94.5P_1{V}_1$
• B. $54.5P_1{V}_1\text{and}94.5P_1{V}_1$
• C. $9P_1{V}_1\text{and}0$
• D. $79.5P_1{V}_1$ and defined $(\because \text{P varies})$

Answer 1

The correct option is A $79.5P_1{V}_1\text{and}94.5P_1{V}_1$

$\omega$ = work under PV curve.
$△E=n{C}_v△T=\frac{P_f{V}_f-P_i{V}_i}{\gamma -1}$
$q=△E-\omega$
$△H=\gamma \times △E=\frac{\gamma }{\gamma -1}(P_f{V}_f-P_i{V}_i)$
$\gamma$ for diatomic $=1+\frac{2}{5}=\frac{7}{5}$
Here, $|\omega |=$ area of trapezium
$=\frac{1}{2}\times 3V_1(P_1+7P_1)=12P_1{V}_1$
Expansion Hence, $\omega =12P_1{V}_1$
$△E=\frac{28P_1{V}_1-P_1{V}_1}{\frac{7}{5}-1}=\frac{5}{2}\times 27P_1{V}_1$
$=67.5P_1{V}_1$
$\therefore q=67.5P_1{V}_{-}(-12P_1{V}_1)$
$=79.5P_1{V}_1$
$△H=\gamma △E=\frac{7}{5}\times \frac{5}{2}\times 27P_1{V}_1=94.5P_1{V}_1$
$\therefore \left[A\right]$